Question: Graph this system of equations and solve. $-3x+4y = -16$ $-4x-4y = -12$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Explanation: Convert the first equation, $-3x+4y = -16$ , to slope-intercept form. $y = \dfrac{3}{4} x - 4$ The y-intercept for the first equation is $-4$ , so the first line must pass through the point $(0, -4)$ The slope for the first equation is $\dfrac{3}{4}$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move up You must also move $4$ positions to the right. $4$ positions to the right. $3$ positions up from $(0, -4)$ is $(4, -1)$ Graph the blue line so it passes through $(0, -4)$ and $(4, -1)$ Convert the second equation, $-4x-4y = -12$ , to slope-intercept form. $y = - x + 3$ The y-intercept for the second equation is $3$ , so the second line must pass through the point $(0, 3)$ The slope for the second equation is $-1$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move down (because it's negative) You must also move $1$ positions to the right. $1$ position to the right. $1$ position down from $(0, 3)$ is $(1, 2)$ Graph the green line so it passes through $(0, 3)$ and $(1, 2)$ The solution is the point where the two lines intersect. The lines intersect at $(4, -1)$.